LearnYMA01WST01 · Statistics 1Discrete random variablesExpectation, variance and the discrete uniform distribution

Discrete random variables · Subtopic 2 of 2

Expectation, Variance and the Discrete Uniform Distribution

$E(X)=\sum xP(X=x)$ is the long-run average. $\text{Var}(X)=E(X^2)-[E(X)]^2$ measures spread. For linear changes, $E(aX+b)=aE(X)+b$ but $\text{Var}(aX+b)=a^2\text{Var}(X)$ : adding a constant shifts values without spreading them.

Edexcel YMA01 WST01AQA 7367 Paper 3≈ 35 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Calculate $E(X)$ , $E(X^2)$ and $\text{Var}(X)$ .
Use $E(aX+b)$ and $\text{Var}(aX+b)$ .
Find unknown probabilities from a given mean.
Use the discrete uniform distribution and its mean and variance.

Pause and think

Everyone in a class gets $5$ bonus marks. What happens to the average mark, and to how spread out the marks are?

Section 2 of 14

Core idea

Start hereThe long-run average

Roll a die thousands of times and average the scores: you'll get very close to 3.5, a number the die can never show. That long-run average is the expectation.

The big idea

$E(X)=\sum xP(X=x)$ ; $\operatorname{Var}(X)=E(X^2)-[E(X)]^2$ ; $E(aX+b)=aE(X)+b$ ; $\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$ .

1. Calculating

E(X)=\sum xp\qquad\operatorname{Var}(X)=\sum x^2p-\left(\sum xp\right)^2

$p$: $P(X=x)$ .

Worked example $x=1,2,3,4$ with $p=0.1,0.3,0.4,0.2$

$E(X)$ , $\operatorname{Var}(X)$ , and for $Y=3X-2$ .

Goal: Four values.
Strategy: Table, then rules.

$E(X)=0.1+0.6+1.2+0.8=2.7$
$\sum xp$ .
$E(X^2)=0.1+1.2+3.6+3.2=8.1$
$\sum x^2p$ .
$\operatorname{Var}(X)=8.1-2.7^2=0.81$
Formula.
$E(Y)=3(2.7)-2=6.1,\ \operatorname{Var}(Y)=9(0.81)=7.29$
Linear rules.

Answer

2.7

0.81

;

6.1

7.29

✓ Check

E(X)

lies between 1 and 4, nearer 3 where probability is largest ✓.

Common misconception $E(X^2)\neq[E(X)]^2$

8.1

7.29

: the difference is the variance.

2. Discrete uniform

Discrete uniform: a fair die

All bars equal. For

1

n

E(X)=\dfrac{n+1}{2}

and

\text{Var}(X)=\dfrac{n^2-1}{12}

. For a die,

\dfrac{35}{12}=2.92

Mean and variance for

1,\ldots,n

$E(X)=\frac1n\cdot\frac{n(n+1)}2=\frac{n+1}2$
Sum of $1$ to $n$ .
$E(X^2)=\frac1n\cdot\frac{n(n+1)(2n+1)}6=\frac{(n+1)(2n+1)}6$
Sum of squares.
$\operatorname{Var}=\frac{(n+1)(2n+1)}6-\frac{(n+1)^2}4=\frac{(n+1)[2(2n+1)-3(n+1)]}{12}$
Common factor $(n+1)$ , denominator 12.
$=\frac{(n+1)(n-1)}{12}=\frac{n^2-1}{12}$
$4n+2-3n-3=n-1$ .

Worked exampleA fair die

$E(X)$ , $\operatorname{Var}(X)$ , $E(2X+1)$ .

Goal: Values.
Strategy: $n=6$ .

$E=\tfrac72=3.5,\ \operatorname{Var}=\tfrac{35}{12}$
Formulas.
$E(2X+1)=8$
Linear rule.

Answer

3.5

\frac{35}{12}

8

✓ CheckDirect:

\frac{1+4+\cdots+36}6=\frac{91}6

;

\frac{91}6-12.25=2.917=\frac{35}{12}

✓.

Summary

Mean and variance from the table.
Linear rules.
Uniform formulas.

Expectation: Long-run mean.
Discrete uniform: All values equally likely.

HistoryHuygens' expectation. Christiaan Huygens wrote the first printed book on probability in 1657 and introduced the idea of the "value of a chance", what we call the expected value: the fair price to pay to play a game of chance.

Fun factExpectation you can never get. The expected score on a fair dice is

3.5

, a value you can never actually roll. Expectation is a long-run average, not a likely single result.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in